a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(P=\dfrac{x^2-4x+3-2x-6-x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-5}{x+3}\)
\(=\dfrac{-6x-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{2x+1}=\dfrac{-3}{x-3}\)
b: Thay x=5 vào P, ta được:
\(P=\dfrac{-3}{5-3}=\dfrac{-3}{2}\)
c: Để P=3/7 thì x-3=-7
hay x=-4(nhận)